જો a = lim_{x rightarrow 0} frac{sqrt{1+sqrt{ | vedclass

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(B) માટે: $a = \lim_{x ightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} 	imes \frac{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+x^4}}+\s

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$32$

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(B) માટે: $a = \lim_{x ightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} 	imes \frac{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}} = \lim_{x ightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})} = \lim_{x ightarrow 0} \frac{x^4}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.$b$ માટે: $b = \lim_{x ightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}} = \lim_{x ightarrow 0} \frac{(1-\cos^2 x)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} = \lim_{x ightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x} = \lim_{x ightarrow 0} (1+\cos x)(\sqrt{2}+\sqrt{1+\cos x}) = (1+1)(\sqrt{2}+\sqrt{2}) = 2 	imes 2\sqrt{2} = 4\sqrt{2}$.$ab^3$ ની ગણતરી: $ab^3 = \frac{1}{4\sqrt{2}} 	imes (4\sqrt{2})^3 = \frac{1}{4\sqrt{2}} 	imes 64 	imes 2\sqrt{2} = \frac{128\sqrt{2}}{4\sqrt{2}} = 32$.

જો $a = \lim_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ અને $b = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ હોય,તો $ab^3$ ની કિંમત શોધો.

Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + \frac{3}{{1 - {n^2}}} + \dots + \frac{n}{{1 - {n^2}}}} \right] =$

જો ${x_n} = \frac{{1 - 2 + 3 - 4 + 5 - 6 + \dots - 2n}}{{\sqrt {{n^2} + 1} + \sqrt {4{n^2} - 1} }},$ હોય,તો $\mathop {\lim }\limits_{n \to \infty } {x_n}$ ની કિંમત શોધો.

જો $f(x) = \begin{cases} x^2-1, & 0 < x < 2 \\ 2x+3, & 2 \leq x < 3 \end{cases}$ હોય,તો તે દ્વિઘાત સમીકરણ શોધો જેના બીજ $\lim_{x \rightarrow 2^{-}} f(x)$ અને $\lim_{x \rightarrow 2^{+}} f(x)$ હોય.

જો $x_1 = 3$ અને $x_{n+1} = \sqrt{2 + x_n}$ હોય,તો $\lim_{n \to \infty} x_n$ ની કિંમત શોધો.

$\lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3} = $

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